Sciencemadness Discussion Board

2NaCl + H2SO4

CHRIS25 - 8-4-2014 at 11:08

Not quite understanding how to understand something.... Seeing that 2 moles of NaCl prevent a Hydrogen atom from bonding with Na2SO4 I wanted to understand why. I looked at the molecular structures of each chemical in the hope of deducing something but I am utterly new to this way of looking at reactions. So what as a complete beginner should I be reading up about in order to understand why 2 Sodium atoms will bond with a sulphate molecule and prevent a hydrogen atom from bonding instead and thereby giving the bisulphate.

Galinstan - 8-4-2014 at 11:43

the sodium IONS don't stop the bisulphate ion from forming but the bisulphate ion is in it's self quite acidic so will dissociate in aquoas solution and then hydrochloic acid can be distilled off leaving you with sodium sulphate.

whetther the sulphate or bisulphate is more to do with stiochometry than anything else in this reaction.

[Edited on 8-4-14 by Galinstan]

vmelkon - 8-4-2014 at 12:23

Quote: Originally posted by CHRIS25  
Not quite understanding how to understand something.... Seeing that 2 moles of NaCl prevent a


I'm assuming you are talking about mixing solid NaCl and high conc H2SO4 and applying heat to drive off the HCl gas.
Almost no reaction is clean.
Most likely, Na2SO4 and NaHSO4 will form. Chemical equations are just idealized representations. In the real word, molecules are moving around at high speed in random directions and need to bump into each other at the proper angle and with sufficient force.

CHRIS25 - 8-4-2014 at 12:27

This does not make sense because you say:.."bisulphate ion is in it's self quite acidic.." I know it is, but it has not formed. When you mix 2NaSO4 and H2SO4 you have essentially a soup of sulphate, Sodium, Hydrogen and Chloride ions in solution. By adding extra Sodium and Chloride (though I see the latter does not influence anything) by adding extra Na ions you stop the Hydrogen from bonding with the sulphate. Now I accept my perspective may be wrong but the observation is that 2 positive Na ions are attaching to the negative O ion whereas with less Na only one Na ion attaches to the negative O ion and a Hydrogen Ion is allowed to bond with an oxygen. (my terminology may well be wrong. But I have only just started to look into this sort of thing so I am new to it).

CHRIS25 - 8-4-2014 at 12:30

Quote: Originally posted by vmelkon  
Quote: Originally posted by CHRIS25  
Not quite understanding how to understand something.... Seeing that 2 moles of NaCl prevent a


I'm assuming you are talking about mixing solid NaCl and high conc H2SO4 and applying heat to drive off the HCl gas.
Almost no reaction is clean.
Most likely, Na2SO4 and NaHSO4 will form. Chemical equations are just idealized representations. In the real word, molecules are moving around at high speed in random directions and need to bump into each other at the proper angle and with sufficient force.

Hi again, Yes I am. But I do not want the bisulphate unfortunately. I assumed by shifting the reaction balanced more to providing extra Na then I would ensure a sulphate result.

gdflp - 8-4-2014 at 15:43

Just out of curiosity, why dont you test your product with NaHCO3. If it bubbles, sodium bisulfate is present. In that case, just add more NaHCO3 until the gas evolution stops, thus giving pure sulfate.

HgDinis25 - 8-4-2014 at 15:52

The reaction that takes place is:
NaCl + H2SO4 --> NaHSO4 + HCl
However, the NaHSO4 formed will also react with NaCl:
NaHSO4 + NaCl --> Na2SO4 + HCl

That's why, the all process is simplifyied to:
2NaCl + H2SO4 --> Na2SO4 + 2HCl
However, during the course of the reaction, you will have all species present. If you had the precise amount, though, (2 moles of NaCl, one mole of H2SO4), you'll end up with only HCl and Na2SO4. If you have an excess of H2SO4 you'll end up with NaHSO4 in excess. If you have an excess of NaCl, it won't react.

CHRIS25 - 8-4-2014 at 23:56

gdfip: I do not have any bisulphate.
HgDinis25: Ah ok then. So An experiment where I first need to add phenolpthalein in order to see when there is precise neutralization of the product on a small scale, then double up all the figures for the actual process. Glad I asked. And thankyou for the explanation.

Decided to use NaOH instead, don't like the heat but at least I can cool, but I really actually don't want HCl fumes, pity because that would make an excellent by-product bubbled through water but I do not have the equipment for this. Find neutral point, keep balanced on NaOH side

[Edited on 9-4-2014 by CHRIS25]

blogfast25 - 9-4-2014 at 04:34

Quote: Originally posted by HgDinis25  
The reaction that takes place is:
NaCl + H2SO4 --> NaHSO4 + HCl
However, the NaHSO4 formed will also react with NaCl:
NaHSO4 + NaCl --> Na2SO4 + HCl



W/o heating it tends to stop at the first displacement. To react bisulphate with NaCl requires quite strong heating.

There are several threads on HCl generators on this forum. 'Search and ye shall find'...

[Edited on 9-4-2014 by blogfast25]

CHRIS25 - 9-4-2014 at 07:44



"...W/o heating it tends to stop at the first displacement. To react bisulphate with NaCl requires quite strong heating".....

Which is why I will use NaOH. But the information is appreciated since I would never have known from the balanced equations that this is what was needed.

[Edited on 9-4-2014 by CHRIS25]

gdflp - 9-4-2014 at 08:00

No, I was saying that any bisulfate produced in the reaction could be tested for and neutralized to sulfate using NaHCO3.

annaandherdad - 9-4-2014 at 08:40

I prefer the NaHSO4 method for making HCl, since it makes me nervous working with flasks full of hot H2SO4. Never had a flask break on me, but I'm not looking forward to the first time, either.

CHRIS25 - 9-4-2014 at 09:01

Quote: Originally posted by gdflp  
No, I was saying that any bisulfate produced in the reaction could be tested for and neutralized to sulfate using NaHCO3.

Sorry but I am finding this a little confusing, if you know you are producing bisulphate in the reaction, why test for it? more to the point I do not at all understand the procedure here in practise. If I have my reaction going, in the middle of it I am supposed to test for bisulphate? Then let it continue ...to what? This really does not make sense at all sorry. Why interfere wit a reaction when you know that by heating it you will produce the sul;phate if the stoichemetry is correct and there is Not more H2SO4 than Na in the soln. Why is this being made so complicated? and you say neutralize the soln by using bisulphate, well the bisulphate is already there is not?
I understand that you may know what you mean, but you may well have condensed a whole set of steps into one sentence that from my perspective I fail to follow into the practical out workings, while at the same time I think I understand the theory.

[Edited on 9-4-2014 by CHRIS25]

[Edited on 9-4-2014 by CHRIS25]

woelen - 9-4-2014 at 10:19

Quote: Originally posted by CHRIS25  
Not quite understanding how to understand something.... Seeing that 2 moles of NaCl prevent a Hydrogen atom from bonding with Na2SO4 I wanted to understand why. I looked at the molecular structures of each chemical in the hope of deducing something but I am utterly new to this way of looking at reactions. So what as a complete beginner should I be reading up about in order to understand why 2 Sodium atoms will bond with a sulphate molecule and prevent a hydrogen atom from bonding instead and thereby giving the bisulphate.

First you must keep in mind that Na(+) ions do not really bind to anything else. They exist as separate entities in the crystal lattice and they also exist as separate entities in solution (with water molecules around them with their more negatively charged part turned towards the positive sodium ions).

A compound like NaCl can better be written as Na(+)Cl(-) and a compound like Na2SO4 can better be written as 2Na(+)[SO4](2-). This compound has SO4(2-) entities in its crystal lattice and Na(+) entities in a ratio of 1 : 2.

Read about the difference between ionic and covalent bonds:

http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/I...

Within the sulfate ion, there is covalent bonding, this entire ion having charge -2. The compound Na2SO4 hence is an ionic bonded compound with one of its entitities itself in turn being a covalently bonded entity.

HCl is covalently bonded, but when it reacts with water, then first some complex with water is formed and this breaks apart in ions:

H2O + HCl --> H2OHCl --> H3O(+) + Cl(-)

The H3O(+) ions and Cl(-) ions start swimming around through the liquid, independent of each other. Even in quite strong hydrochloric acid the separation in ions is nearly complete, but in 35% HCl, part of the molecules of HCl still exists as covalent molecules. These can escape from the liquid and then react with water vapor from the air, giving ions and small droplets of solutions of these ions. This is the fuming of concentrated HCl.

H2SO4 is a completely covalently bonded species, O2S(OH)2, but one of the hydrogen ions very easily splits off in the presence of water, but also in the presence of other ions, e.g.

O2S(OH)2 + Cl(-) <--> [O2S(OH)OHCl](-) <--> [O2S(OH)(O)](-) + HCl.

Because HCl is a gas, this escapes and the reaction hence is driven to the right.

The H of the species [O2S(OH)(O)](-) (which is bisulfate ion) is more tightly bonded to its oxygen and more effort has to be put in removing this as well. When one hydrogen is split off, together with a +1 charge, then a charge shift occurs in the remaining covalent structure and the second hydrogen atom becomes more tightly bonded because of this.

CHRIS25 - 9-4-2014 at 10:43

Woelen thanks, I understood everything up until this point: ....When one hydrogen is split off, together with a +1 charge, then a charge shift occurs in the remaining covalent structure and the second hydrogen atom becomes more tightly bonded because of this. ......
I am looking at the covalent structure of the Bisulphate and see one H+ ion bonded to one O ion. But then you said that when this breaks off then the charge shift occurs? and another H+ ion joins it? Do not quite follow.
Do you mean by this that everytime a H+ ion breaks away from the Bisulphate to become a sulphate that then another H+ ion is running to takes its place and make it a bisulphate again?
[Edited on 9-4-2014 by CHRIS25]

[Edited on 9-4-2014 by CHRIS25]

blogfast25 - 10-4-2014 at 05:21

Quote: Originally posted by CHRIS25  
Do not quite follow.

[Edited on 9-4-2014 by CHRIS25]


I'll make it real simple for you. The bisulphate ion is negatively charged, so 'ripping out' a positively charged particle like a proton means that the electrostatic attraction has to be overcome. This is not the base for H2SO4 === > H+ + HSO4(-), so sulphuric acid is a stronger acid than HSO4(-).

This is a general rule for multiprotonic acids: H3PO4 > H2PO4(-) > HPO4(2-), H2CO3 > HCO3(-) etc etc. Each time the negatively charged species is a weaker acid because of the extra energy needed to rip the proton away from it. So as a general rule: Ka1 > Ka2 > Ka3 etc.

[Edited on 10-4-2014 by blogfast25]

CHRIS25 - 10-4-2014 at 06:44

Thankyou blogfast,


I neutralised a 7mL test sample of 2M Sulphuric with 2M NaOH from the actual Solution. Then used the calculations to do the full neutralisation. Took the whole solution to Ph12 then brought it back with Sulphuric to PH 8.1

This is not right? I calculate that at least 160mLs of water must evaporate before any precipitate and this should be white. After only 50mLs of water evaporating I have this precipitate. It does not look like Sodium sulphate does it. Any ideas. I have considered contamination but do not know how really.

IMG_1334.JPG - 78kB

blogfast25 - 10-4-2014 at 08:28

C25:

I'm not sure what it is you're trying to prepare but I'm guessing sodium sulphate from NaOH and H2SO4:

2 NaOH + H2SO4 === > Na2SO4 + 2 H2O

When you're at pH 8.1, you're very close to 100 % neutralisation (there's not much point in going any further but just another drop of H2SO4 solution will probably get you there).

Your 'floaters' look suspiciously like Fe(OH)3.nH2O. Iron is a common contaminant in beginners' concoctions: it may have come from the sulphuric acid, from the NaOH or from the reactor.

Simply filter off the brown bits and you should have a solution of technical Na2SO4. Evaporate water to get the solid salt.

When neutralising acids with bases (or vice versa) using pH, be aware of how pH actually varies with added base/acid: close to the neutrality point it's very easy to 'overshoot', look up 'titration curve':

http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.ht...

So one minute you've got pH, say 5 and you're doing OK, next you add just that liiiittle bit too much base and bang, you're at pH 11! Very confusing if you don't know what's going on...

[Edited on 10-4-2014 by blogfast25]

[Edited on 10-4-2014 by blogfast25]

CHRIS25 - 10-4-2014 at 09:14

Hallo Blogfast well I did filter it off after posting, Iron, that is interesting. I have a nice sodium sulphate and hopefully will get it to anhydrous. Now to do a bigger batch. I had no problems at all with exothermic 'overloads', I reckon I can push the concentration up to 3.5M this time and keep the ice bath close just in case. Though i had no need of it at all today. Thankyou for the titration curves, I know what you mean, in my first test I had syringed in 16.4mLs or so of NaOH and it was at a PH of 4-5, just 0.3 or 0.4mLs more and it shot to PH 12+ (I say 12+ because I can not measure beyond 12).
Your help very much appreciated, considering where I started when I first joined the forum (No Knowledge at all) I have come a long way, and thanks to many contributors including yourself.

blogfast25 - 11-4-2014 at 05:18

Quote: Originally posted by CHRIS25  
Hallo Blogfast well I did filter it off after posting, Iron, that is interesting. I have a nice sodium sulphate and hopefully will get it to anhydrous. Now to do a bigger batch. I had no problems at all with exothermic 'overloads', I reckon I can push the concentration up to 3.5M this time and keep the ice bath close just in case.


An ice bath won't save you from excessive exotherms (over boiling or splattering) because cooling takes time. Instead calculate the expected temperature increase, using the Enthalpy of Neutralisation for H3O+ + OH- == 2 H2O and ΔH = m Cp ΔT.

This way you can easily calculate the safe highest concentration of acid and NaOH to be used, without any boiling occurring (ΔT < 80 C). But ALWAYS mix the two SLOWLY with intense STIRRING, to avoid local overheating.

[Edited on 11-4-2014 by blogfast25]

CHRIS25 - 11-4-2014 at 06:02

I just used 5M solutions to make another batch. expected yield is 1Mole Na2SO4 anhydrous. My last yield was 15.5g, theoretical yield was 17.3g, so not bad at all. The heat is no problem at all, just a couple of ice packs around everything and it all goes smoothly. I am sorry, but the maths and whole subject of Enthalpy of Neutralisation is way beyond my head and my time at this stage, maybe something for much later on, but not now, just sufficient to know when to be sensible with exothermic reactions. But I take your point and will look into this at some other date. Thankyou for pointing yet another subject of interest to learn.



[Edited on 11-4-2014 by CHRIS25]